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Figure 1 shows the speed-time graph of a cyclist moving on a straight road over a 7 s period - Edexcel - A-Level Maths Mechanics - Question 1 - 2006 - Paper 1

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Figure 1 shows the speed-time graph of a cyclist moving on a straight road over a 7 s period. The sections of the graph from t = 0 to t = 3, and from t = 3 to t = 7,... show full transcript

Worked Solution & Example Answer:Figure 1 shows the speed-time graph of a cyclist moving on a straight road over a 7 s period - Edexcel - A-Level Maths Mechanics - Question 1 - 2006 - Paper 1

Step 1

a) the graph from t = 0 to t = 3 is a straight line.

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Answer

The graph being a straight line indicates that the velocity of the cyclist is changing at a constant rate. This implies that the cyclist is experiencing constant acceleration.

Step 2

b) the graph from t = 3 to t = 7 is parallel to the t-axis.

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Since the graph is parallel to the t-axis, it indicates that the cyclist is moving with a constant speed during this interval. The speed does not change, meaning the cyclist is not accelerating or decelerating.

Step 3

c) Find the distance travelled by the cyclist during this 7 s period.

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Answer

To calculate the distance travelled, we can analyze the graph. The area under the speed-time graph represents the distance. From t = 0 to t = 3, the distance can be calculated using the area of a triangle:

Area=12×base×height=12×3×5=7.5 m\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 5 = 7.5 \text{ m}

From t = 3 to t = 7, the distance can be calculated using the area of a rectangle:

Area=length×width=4×5=20 m\text{Area} = \text{length} \times \text{width} = 4 \times 5 = 20 \text{ m}

Now, adding both areas together gives us the total distance:

Total Distance=7.5extm+20extm=27.5extm\text{Total Distance} = 7.5 ext{ m} + 20 ext{ m} = 27.5 ext{ m}

Thus, the total distance travelled by the cyclist during the 7 s period is 27.5 m.

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